Question on finding the volume of vapor and the volume of liquid

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SUMMARY

The discussion focuses on calculating the volume of evaporated water and comparing it to the volume of liquid water during perspiration on a hot day. The user successfully applies the ideal gas law (PV=nRT) to determine the volume of vapor produced from 1.0 kg of water, resulting in a volume of 1.402 m³. Key concepts discussed include vapor pressure and the density of water vapor, with the user noting the importance of atmospheric pressure in their calculations.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of vapor pressure and its significance in phase changes
  • Familiarity with the density of water and water vapor
  • Basic concepts of thermodynamics related to phase diagrams
NEXT STEPS
  • Research the concept of vapor pressure and its tables for various temperatures
  • Learn about the ideal gas law applications in real-world scenarios
  • Explore the phase diagram of water and its implications on state changes
  • Study the properties of gases, including density and molar mass calculations
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and anyone interested in understanding the principles of phase changes and gas laws in practical applications.

rinchan4
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Homework Statement



On a hot 35 degrees celsius day you perspire 1.0 kg of water during your workout.
A) what volume is occupied by the evaporated water?
B) By what factor is this larger than the volume occupied by the liquid water?

Homework Equations



PV=nRT
PV=NKbT
density of water is 1000 kg/m3= M/V
n=Mass/ Molar mass


3. The Attempt at a Solution [/b

I really have no idea on how to start this problem. I did solve for n=M/Mmol=1000g/18.02g/mol= 55.49 moles but i don't even know if I need it. Any help would be greatly appreciated.
 
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I just need a hint to get going in the right direction. I'm thinking that this problem may have something to do with the phase diagram of water?
 
Welcome to PF.

What's missing here is the pressure of water. Has your class or book discussed the concept of vapor pressure ? There might be a table of water's vapor pressure in your book, for different temperatures. That information is needed to solve this.
 
Yeah we didn't talk at all about vapor pressure and the only thing that it says in my book is that we can ignore vapor pressure if we keep the tempature low. I did find that the vapor pressure of water is 2.4 kPa and 20 degrees celsius.
 
I tried using the density of vapor 0.0022(since i knew the mass of 1kg) and from that find the volume of 454.5m3 but that wasn't right
 
You had pV=nRT and you had both n and T, so all you need is p. Imagine trapping all of the water vapor in a box that's free to expand or contract whenever it needs to. What would the pressure of the water vapor be?
 
would be the same as the atmospheric pressure of 101300 Pa?
 
Yeah.
 
so then...
P= 101300 Pa
n= 1000/18.02=55.49moles
T= 308 k
R=8.31
Solve for V and I get 1.402 m3
 
  • #10
awesome thank you so much. I was able to figure out part b. Thank you! :)
 

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